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  • If <img alt="\mathrm{A}\left(1, \mathrm{p}^{2}\right) ; \mathrm{B}(0,1)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%7D%5Cleft%281%2C%20%5Cmathrm%7Bp%7D%5E%7B2%7D%5Cright%29%

If \mathrm{A}\left(1, \mathrm{p}^{2}\right) ; \mathrm{B}(0,1) and \mathrm{C}(\mathrm{p}, 0) are the coordinates of three points then the value of \mathrm{p}  for which the area of the triangle \mathrm{ABC} is minimum, is

Option: 1

\frac{1}{\sqrt{3}}


Option: 2

-\frac{1}{\sqrt{3}}


Option: 3

\frac{1}{\sqrt{3}}$ or $-\frac{1}{\sqrt{3}}


Option: 4

none


Answers (1)

best_answer

\mathrm{A=\frac{1}{2}\left|\begin{array}{ccc}1 & p^{2} & 1 \\ 0 & 1 & 1 \\ p & 0 & 1\end{array}\right|=\frac{1}{2}\left[1(1-0)+p\left(p^{2}-1\right)\right]=\frac{1}{2}\left(p^{3}-p+1\right)}

\mathrm{Hence \: A=\frac{1}{2}\left|p^{3}-p+1\right|}

Now, minimum value of modulus is zero. Since \mathrm{A(p)} is a cubic it must vanish for some \mathrm{p} other than given as \mathrm{\mathrm{A}, \mathrm{B}, \mathrm{C} \Rightarrow (D) }

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